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l2ans (1)

# l2ans (1) - Outline Conditional Probability Conditional...

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Unformatted text preview: Outline Conditional Probability Sept 21, 2010 Conditional Probability Outline Outline 1 Conditional Probability 2 Bayes Theorem 3 Independence Conditional Probability Conditional Probability Bayes Theorem Independence Outline 1 Conditional Probability 2 Bayes Theorem 3 Independence Conditional Probability Conditional Probability Bayes Theorem Independence Overview Conditional probabilities are difficult for people to deal with We need a formal mathematical framework What do we mean by conditional probability? Say we care about the probability that a patient has prostate cancer: Pr ( PrCa ) What if the patient has a positive PSA screening? Pr ( PrCA | PSA +) The probability he has prostate cancer depends (is conditioned on) the PSA test Conditional Probability Conditional Probability Bayes Theorem Independence Tossing Die Consider tossing a die S = { 1 , 2 , 3 , 4 , 5 , 6 } Let the event E 1 =a number > 3. E 1 = { 4 , 5 , 6 } Let the event E 2 =an even number. E 2 = { 2 , 4 , 6 } Pr ( E 1 ) = 1 / 2 Pr ( E 2 ) = 1 / 2 What is the probability of E 1 if we know the number is even? By saying the number is even, we’ve restricted the sample space to S r = { 2 , 4 , 6 } Need to redefine E 1 , r = { 4 , 6 } Pr ( E 1 if E 2 is true ) = | E 1 , r | | Sr | = 2 3 Conditional Probability Conditional Probability Bayes Theorem Independence Def’n Conditional Probability Fortunately, we don’t need to formally restrict the sample space each time Definition Pr ( A | B ) = Pr ( A ∩ B ) Pr ( B ) Pr ( A | B ) is a probability function that must follow the standard axioms and properties discussed last lecture For some event B , where Pr ( B ) > 1 Pr ( A | B ) ≥ 2 P ( S | B ) = 1 3 if A 1 , A 2 are mutually exclusive given C, then Pr ( A 1 ∪ A 2 | C ) = Pr ( A 1 | C ) + Pr ( A 2 | C ) 4 Pr ( ∅| C ) = 5 etc Conditional Probability Conditional Probability Bayes Theorem Independence Restricting the Sample Space Conditional Probability Conditional Probability Bayes Theorem Independence How do we justify this, formally? S = { s 1 , s 2 , s 3 , . . . } We can attach a probability to each element in the sample space, Pr ( s j ) Now if we condition on the event E , we want Pr ( s j | E ) = , ∀ s j ∈ E c And we want to define some new probability function f ( s j | E ) ∀ s j ∈ E We should have the same relative magnitude for the probabilities before and after conditioning f ( s j | E ) = cPr ( s j ) , ∀ s j ∈ E We also need ∑ s j ∈ E f ( s j | E ) = 1 = ∑ s j ∈ E cPr ( s j ) Thus c = 1 / ∑ s j ∈ E Pr ( s j ) = 1 / Pr ( E ) Finally, Pr ( s j | E ) = Pr ( s j ) / Pr ( E ) Conditional Probability Conditional Probability Bayes Theorem Independence Example Riedler et al (Lancet, 2001) conducted a cross-sectional survey of the the rural areas of Austria, Germany and Switzerland. The study estimated the relationship between exposure to an agricultural environment in early childhood (namely exposure to stables and farm milk in the 1st year of life) and a diagnosis of asthma.milk in the 1st year of life) and a diagnosis of asthma....
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l2ans (1) - Outline Conditional Probability Conditional...

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